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kupik [55]
3 years ago
5

An equipoterntial surface that surrounds a + 3.0 pC point charge has a radius of 2.0 cm. What is the potential of this surface?​

Physics
1 answer:
mestny [16]3 years ago
8 0

Answer:

Electric potential = 0.00054 V

Explanation:

We are given;

Charge; q = 3 pC = 3 × 10^(-12) C

Radius; r = 2 cm = 0.02 m

Formula for the electric potential of this surface will be;

V = kqr

Where;

K is a constant = 9 × 10^(9) N⋅m²/C².

Thus;

V = 9 × 10^(9) × 3 × 10^(-12) × 0.02

V = 0.00054 V

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Andreas93 [3]

Answer:

28Nm

Explanation:

Torque is expressed as the prduct of force and radius

Given

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Torque = 140 * 0.2

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Torque = 28Nm

Hence the amouunt of torque applied is 28Nm

8 0
3 years ago
Jake drove 160 kilometres in 2 hours. What was his speed, in kmh-1
ruslelena [56]

Answer:

80 kmh

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5 0
3 years ago
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aleksklad [387]

Answer:

1.23×10⁸ m

Explanation:

Acceleration due to gravity is:

a = GM / r²

where G is the universal gravitational constant,

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When the object is on the surface of the Earth, a = g and r = R.

g = GM / R²

When the object is at height i above the surface, a = 1/410 g and r = i + R.

1/410 g = GM / (i + R)²

Divide the first equation by the second:

g / (1/410 g) = (GM / R²) / (GM / (i + R)²)

410 = (i + R)² / R²

410 R² = (i + R)²

410 R² = i² + 2iR + R²

0 = i² + 2iR − 409R²

Solve with quadratic formula:

i = [ -2R ± √((2R)² − 4(1)(-409R²)) ] / 2(1)

i = [ -2R ± √(1640R²) ] / 2

i = (-2R ± 2R√410) / 2

i = -R ± R√410

i = (-1 ± √410) R

Since i > 0:

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R = 6.37×10⁶ m:

i ≈ 1.23×10⁸ m

8 0
3 years ago
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3 years ago
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A meteor moving 468 km per minute traveling in a south-to-north direction passed near Earth in 2013. Does this statement describ
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Answer:

this statement describes meteor's velocity,

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